Question #103626
A biquadratic equation must have at least one real root. Is it true or false? give reasons.
1
Expert's answer
2020-03-02T09:23:17-0500

A biquadratic equation is a polynomial equation with degree 4 without having degree 1 and 3 terms. Since it is saying it "must have" at least one real root, it makes the statement false

To prove this I will provide a counter example.

Consider a biquadratic equation

x4+7x2+12=0x^4+7x^2+12=0

For finding roots , let u=x2u=x^2

Given equation will rewritten as

u2+7u+12=0u^2+7u+12=0

u2+3u+4u+12=0u^2+3u+4u+12=0

u(u+3)+4(u+3)=0u(u+3)+4(u+3)=0

(u+3)(u+4)=0(u+3)(u+4)=0

i.eu=3i.e u=-3 or u=4u=-4

sincesince u=x2u=x^2

x2=3x^2=-3 or x2=4x^2=-4

x=±3x=±\sqrt{\smash[b]{-3}} x=±3i±\sqrt{\smash[b]{3}}i or x=±4x=±\sqrt{\smash[b]{-4}} or x=±2ix=±2i


Which means a biquadratic equation can have all imaginary roots.



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